It takes 0.333 mol (33.7 g) of KNO₃ to make 21 g of HNO₃.
KNO₃ reacts with concentrated H₂SO₄ to produce HNO₃ and KHSO₄.
[tex]\text{KNO}_3 + \text{H}_2\text{SO}_4 \; (\text{conc.}) \to \text{HNO}_3 + \text{KHSO}_4[/tex] (balanced.)
Each mole formula unit of KNO₃ reacts with excess H₂SO₄ to produce one mole of HNO₃. In other words, it takes the same number formula units of KNO₃ to produce a certain number of moles of HNO₃.
How many moles of molecules in 21 g of HNO₃?
Relative atomic mass:
Molar mass of HNO₃: [tex]1.008 + 14.007 + 3 \times 15.999 = 63.01 \; \text{g}\cdot \text{mol}^{-1}[/tex].
[tex]n(\text{HNO}_3) = \dfrac{m(\text{HNO}_3)}{M(\text{HNO}_3)} = \dfrac{21\; \text{g}}{63.012\; \text{g}\cdot \text{mol}^{-1}} = 0.333 \; \text{mol}[/tex].
It takes 0.333 moles formula units of KNO₃ to produce 21 g or 0.333 moles of HNO₃.
What's the mass of 0.333 moles formula units of KNO₃?
Molar mass of KNO₃: [tex]39.098 + 14.007 + 3 \times 15.999 = 101.10 \; \text{g}\cdot \text{mol}^{-1}[/tex]
[tex]m(\text{KNO}_3) = n(\text{KNO}_3) \cdot M(\text{KNO}_3)\\\phantom{m(\text{KNO}_3)} = 0.333 \; \text{mol} \times 101.10 \; \text{g}\cdot \text{mol}^{-1}\\\phantom{m(\text{KNO}_3)} = 33.7 \; \text{g}[/tex].
